What volume of water can be boiled by 3.0 kJ of energy? (Refer to table of constants for water.)A.3.0 kJ × × 18.02 g/mol × = 13 mLB.3.0 kJ × × 18.02 g/mol × = 1.3 mLC.3.0 kJ × × 18.02 g/mol × = 9.0 mLD.3.0 kJ × × 18.02 g/mol × = 0.19 mL

To solve this problem, we need to use the specific heat capacity of water and the formula for calculating heat energy. The specific heat capacity of water is 4.18 J/g°C and the heat of vaporization of water is 40.7 kJ/mol.

The formula for calculating heat energy is Q = m * C * ΔT, where Q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature. However, since we are boiling the water, we need to use the heat of vaporization instead of the specific heat capacity.

The formula becomes Q = m * Hv, where Hv is the heat of vaporization.

We can rearrange the formula to solve for m (mass): m = Q / Hv

First, we need to convert the energy from kJ to J: 3.0 kJ = 3000 J

Then, we need to convert the heat of vaporization from kJ/mol to J/g: 40.7 kJ/mol = 40.7 J/g (since the molar mass of water is approximately 18.02 g/mol)

Now we can plug in the values into the formula: m = 3000 J / 40.7 J/g = 73.7 g

Finally, we need to convert the mass to volume. Since the density of water is approximately 1 g/mL, the volume of water that can be boiled by 3.0 kJ of energy is approximately 73.7 mL.

Therefore, none of the provided options (A, B, C, D) are correct.